1. CS654: Digital Image Analysis
Lecture 14: Properties of DFT

2. Recap of Lecture 13 Introduction to DFT 1D and 2D DFT - Unitary
Separability of DFT
Computational complexity
Improvement in computational complexity

3. Outline of Lecture 14 Properties of DFT Translation Periodicity
Conjugate symmetry
Distributivity
Scaling
Average value
Convolution

4. Numerical example 𝐴= 1 2 1 1 1 1 1 −𝑗 −1 𝑗 1 −1 1 −1 1 𝑗 −1 −𝑗
A 4×4 DFT transformation matrix can be written as
𝐴= −𝑗 −1 𝑗 1 −1 1 −1 1 𝑗 −1 −𝑗 𝑢=
𝑽=𝑨𝑼𝑨
𝑣= 1 −1 1 −

5. Translation property of DFT
Let, the input image 𝑢 𝑚,𝑛 is translated to a location 𝑢(𝑚 − 𝑚 0 ,𝑛− 𝑛 0 )
𝑣 𝑘,𝑙 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋(𝑘𝑚+𝑛𝑙) 𝑁
𝑣 𝑡 𝑘,𝑙 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋(𝑘(𝑚− 𝑚 0 )+𝑙(𝑛− 𝑛 0 )) 𝑁
= 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋(𝑘𝑚+𝑙𝑛) 𝑁 exp⁡ 𝑗2𝜋(𝑘 𝑚 0 + 𝑛 0 𝑙) 𝑁
𝑣 𝑡 𝑘,𝑙 =𝑣(𝑘,𝑙)exp⁡ 𝑗2𝜋(𝑘 𝑚 0 + 𝑛 0 𝑙) 𝑁

6. Translation property 𝑣 𝑡 𝑘,𝑙 =𝑣(𝑘,𝑙)exp⁡ 𝑗2𝜋(𝑘 𝑚 0 + 𝑛 0 𝑙) 𝑁
DFT of translated image
𝑣 𝑘− 𝑘 0 ,𝑙− 𝑙 0 =𝑢(𝑚,𝑛)exp⁡ 𝑗2𝜋( 𝑘 0 𝑚+ 𝑙 0 𝑛) 𝑁
Inverse DFT
𝑢(𝑚,𝑛)exp⁡ 𝑗2𝜋 𝑘 0 𝑚+ 𝑙 0 𝑛 𝑁 ↔𝑣(𝑘− 𝑘 0 ,𝑙− 𝑙 0 )
𝑢(𝑚− 𝑚 0 ,𝑛− 𝑛 0 )↔𝑣(𝑘,𝑙)exp⁡ 𝑗2𝜋(𝑘 𝑚 0 + 𝑛 0 𝑙) 𝑁

7. Periodicity 𝑣 𝑘,𝑙 =𝑣 𝑚+𝑁,𝑛 =𝑣 𝑚,𝑛+𝑁 =𝑣(𝑚+𝑛,𝑛+𝑁) ∀𝑘,𝑙
𝑣 𝑘,𝑙 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋(𝑘𝑚+𝑛𝑙) 𝑁
𝑣 𝑘+𝑁,𝑙+𝑁 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋 𝑁 (𝑘+𝑁)𝑚+ 𝑙+𝑁 𝑛
= 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋(𝑘𝑚+𝑛𝑙) 𝑁 𝑒𝑥𝑝 −𝑗2𝜋(𝑥+𝑦)

8. Conjugate symmetry 𝑣 𝑁 2 ±𝑘, 𝑁 2 ±𝑙 = 𝑣 ∗ 𝑁 2 ∓𝑘, 𝑁 2 ∓𝑙 0≤𝑘,𝑙≤ 𝑁 2 −1
When 𝑢(𝑚,𝑛) is real
𝑣 𝑁 2 ±𝑘, 𝑁 2 ±𝑙 = 𝑣 ∗ 𝑁 2 ∓𝑘, 𝑁 2 ∓𝑙
0≤𝑘,𝑙≤ 𝑁 2 −1
𝑘=0 𝑁
𝑁 2
1-D Example
𝑁 2 , 𝑁 2
2-D Example

9. Conjugate symmetry
𝑣 𝑘,𝑙 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋(𝑘𝑚+𝑛𝑙) 𝑁
𝑘= 𝑁 2 ±𝑘,𝑙= 𝑁 2 ±𝑙
𝑣 𝑘,𝑙 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋 𝑁 2 +𝑘 𝑚+ 𝑁 2 +𝑙 𝑛 𝑁

10. Distributivity Scaling
DFT of sum of two signals is equal to the sum of their individual summations
𝐷𝐹𝑇 𝑢 1 𝑚,𝑛 + 𝑢 2 (𝑚,𝑛) =𝐷𝐹𝑇 𝑢 1 (𝑚,𝑛) +𝐷𝐹𝑇 𝑢 2 (𝑚,𝑛)
Scaling
𝑢(𝑎𝑚,𝑏𝑛)↔ 1 |𝑎𝑏| 𝑣 𝑘 𝑎 , 𝑙 𝑏
𝑎,𝑏 are scaling parameters

11. Average value 𝑢 𝑚,𝑛 = 1 𝑁 2 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢(𝑚,𝑛)
𝑢 𝑚,𝑛 = 1 𝑁 2 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢(𝑚,𝑛)
Average value of image
𝑣 𝑘,𝑙 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋(𝑘𝑚+𝑛𝑙) 𝑁
For 𝑘=0,𝑙=0
𝑣 0,0 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 =𝑁 𝑢 𝑚,𝑛
DC Component of an image

12. Rotation 𝑢 𝑟,𝜃 ↔𝑣(𝜔,𝜙) 𝑢 𝑟,𝜃+ 𝜃 0 ↔𝑣(𝜔,𝜙+ 𝜃 0 ) 𝑚=𝑟𝑐𝑜𝑠θ 𝑛=𝑟𝑠𝑖𝑛θ
Polar coordinate in source domain
Instead of working in the Cartesian coordinate, we are working in the polar coordinate
𝑘=𝜔𝑐𝑜𝑠𝜙
𝑛=𝜔𝑠𝑖𝑛𝜙
Polar coordinate in target domain
If we have
𝑢 𝑟,𝜃 ↔𝑣(𝜔,𝜙)
𝑢 𝑟,𝜃+ 𝜃 0 ↔𝑣(𝜔,𝜙+ 𝜃 0 )
Then,

13. Convolution Let there be two images of different size 𝑢 2 𝑚,𝑛 =
(0,0)
(𝑁−1) ≠𝟎
(𝑀−1)
𝑢 1 (𝑚,𝑛)
ℎ(𝑚,𝑛) 𝑚 𝑛
𝒉,𝒎 =𝟎
𝑢 2 𝑚,𝑛 =
𝑚 ′ =0 𝑁−1 𝑛 ′ =0 𝑁−1 ℎ 𝑚− 𝑚 ′ ,𝑛− 𝑛 ′ 𝑐 𝑢 1 𝑚 ′ , 𝑛 ′ ; 0≤𝑚,𝑛≤𝑁−1
ℎ 𝑚,𝑛 𝑐 =ℎ(𝑚 𝑚𝑜𝑑 𝑁,𝑛 𝑚𝑜𝑑 𝑁)

14. Circular Symmetry 𝒉 𝒎− 𝒎 ′ ,𝒏− 𝒏 ′ 𝒄 𝑢 2 𝑚,𝑛 =
(0,0)
(𝑁−1) ≠𝟎
(𝑀−1)
𝑢 1 (𝑚,𝑛)
ℎ(𝑚,𝑛) 𝑚 𝑛
𝒉,𝒎 =𝟎
𝒉 𝒎− 𝒎 ′ ,𝒏− 𝒏 ′ 𝒄 𝑚′ 𝑛′
𝑢 2 𝑚,𝑛 =
𝑚 ′ =0 𝑁−1 𝑛 ′ =0 𝑁−1 ℎ 𝑚− 𝑚 ′ ,𝑛− 𝑛 ′ 𝑐 𝑢 1 𝑚 ′ , 𝑛 ′ ; 0≤𝑚,𝑛≤𝑁−1
𝑢 1 (𝑚′,𝑛′)
Computational complexity??

15. 2D DFT for h in frequency domain
𝐷𝐹𝑇 ℎ 𝑚− 𝑚 ′ ,𝑛− 𝑛 ′ 𝑐 =
𝑚=0 𝑁−1 𝑛=0 𝑁−1 ℎ 𝑚− 𝑚 ′ ,𝑛− 𝑛 ′ 𝑐 𝑊 𝑁 𝑚𝑘+𝑛𝑙
= 𝑊 𝑁 𝑚 ′ 𝑘+ 𝑛 ′ 𝑙 𝑚=0 𝑁−1 𝑛=0 𝑁−1 ℎ 𝑚− 𝑚 ′ ,𝑛− 𝑛 ′ 𝑐 𝑊 𝑁 (𝑚− 𝑚 ′ )𝑘+(𝑛− 𝑛 ′ )𝑙
Let, 𝑝=𝑚−𝑚′ and 𝑞=𝑛−𝑛′
= 𝑊 𝑁 𝑚 ′ 𝑘+ 𝑛 ′ 𝑙 𝑝=−𝑚′ 𝑁−1−𝑚′ 𝑞=−𝑛′ 𝑁−1−𝑛′ ℎ 𝑝,𝑞 𝑐 𝑊 𝑁 𝑝𝑘+𝑞𝑙
= 𝑊 𝑁 𝑚 ′ 𝑘+ 𝑛 ′ 𝑙 𝑝=0 𝑁−1 𝑞=0 𝑁−1 ℎ 𝑝,𝑞 𝑐 𝑊 𝑁 𝑝𝑘+𝑞𝑙

16. DFT of Convolution Function
𝐷𝐹𝑇 𝑢 2 𝑚,𝑛 𝑁 =𝐷𝐹𝑇 ℎ 𝑚,𝑛 𝑁 𝐷𝐹𝑇{ 𝑢 1 (𝑚,𝑛)}
DFT of a two dimensional circular convolution of two arrays is the product of their DFTs

17. Thank you
Next Lecture: Hadamard Transform